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I ask the same question here:https://math.stackexchange.com/q/1019404/192097

writing a little better the previous question: it´s true that if we let $a$ and $b$ be coprime integers, then the arithmetic progesion : $a + bh: h\in Z$, contains a sequence of $k$ "consecutive" primes : $a + nb,a + (n+1)b,...,a + (n+k-1)b$, for possibly all integer $k$?.

I wrote "possibly for all $k$ " because there are some $k$ for which the elements of the progression can not be prime, for example if $a>1$ and if $k>a$ then at least one of $n,(n+1),...,(n+k-1)$ is a multiple of $a$. On the other hand, in the case $a=1$, $b>2$ letting $h=b-2$ is immediate that $1+bh=1+b(b-2)$ is not prime (if $b=3$ then one of: $1+3n$, $1+3(n+1)$ is even), so $k$ is bounded by $b$, whereas in the case $a=1$ and $b=1$: one of $1+n,1+(n+1)$ is even, therefore $k\leq1$. If we require that $a$ and $b$ are large enough, then is not immediate that $k$ need to be small.

informally: in each arithmetic progression, there are "arbitrarily" long sequences of primes.

This question comes after read the Green- Tao theorem on arithmetic progressions but i understand this as: there exist arithmetic progressions of primes, with k terms, where k can be any natural number, wich essentially is not the same as the question before, even more: Green- Tao is a consecuence of the previous one.

summarizing questions:

  • Is this a conjecture?
  • can be found references about the problem in order to try solve it?

I apologize for my poor english. I just speak Spanish. So please excuse me if occasionally the translation is not perfect.

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    $\begingroup$ This has been conjectured but is open. In various generalities: Hardy-Littlewood, Schinzel hypothesis H, Bateman-Horn,... $\endgroup$
    – Felipe Voloch
    Commented Nov 13, 2014 at 21:44
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    $\begingroup$ Specifically if you require every prime $p$ at most $k$ divides $b$, you should be good by Hardy-Littlewood. $\endgroup$
    – Will Sawin
    Commented Nov 13, 2014 at 23:10

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If you meant "consecutive" in the sense that the primes must be consecutive terms in an arithmetic progression, then the conjectures mentioned in the comments have you covered.

If you meant "consecutive" in the sense that the primes are consecutive prime numbers which happen to lie in the same arithmetic progression, then this is actually known to be true.

In 2000, Shiu proved that if $(a,q)=1$, then there are arbitrarily long strings of consecutive primes all equivalent to $a$ modulo $q$. See: http://jlms.oxfordjournals.org/content/61/2/359

The recent work of Maynard and Tao on the problems mentioned in the comments leads to a shorter proof of Shiu's result with the additional conclusion that the consecutive primes must lie in a bounded interval: http://arxiv.org/pdf/1311.5319.pdf

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